S = 1x2 + 2x3 + 3x4 + ... + 38x39 + 39x40

3S = 1x2x3 + 2x3x3 + 3x4x3 + ... + 38x39x3 + 39x40x3

3S = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + ... + 38x39x(40-37) + 39x40x(41-38)

3S = 1x2x3 + 2x3x4-1x2x3 + 3x4x5-2x3x4 + ... + 38x39x40-37x38x39 + 39x40x41-38x39x40

S = 39x40x41 : 3

S = 21320

\(3S=1.2.3+2.3.3+...+39.40.3\)

\(3S=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+39.40.\left(41-38\right)\)

\(3S=0.1.2-1.2.3+1.2.3-2.3.4+...+38.39.40-39.40.41\)

\(3S=30.40.41\)

\(S=10.40.41\)

<span class="label label-info"><i class="fa fa-exclamation-triangle fa-2"></i> Quản lý</span>

Ta có: S = 1 x 2 + 2 x 3 + 3 x 4 +...+ 38 x 39 + 39 x 40

=> 3S = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + ..... + 39 x 40 x (41 - 38)

=> 3S = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + ..... + 39 x 40 x 41

=> 3S = 39 x 40 x 41

=> S = 39 x 40 x 41 : 3

=> S = 21320

Người đầu tiên trả lời nek

Ta có :

\(S=1.2+2.3+...+39.40\)

\(\Rightarrow3S=1.2.3+2.3.3+....+39.40.3\)

\(\Rightarrow3S=1.2.\left(3-0\right)+2.3.\left(4-1\right)+....+39.40.\left(41-38\right)\)

\(\Rightarrow3S=1.2.3-0.1.2+2.3.4-1.2.3+.....+39.40.41-38.39.40\)

\(\Rightarrow3S=39.40.41\)

\(\Rightarrow S=\frac{38.39.40}{3}=21320\)

TL

 S= ( 1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)

3.S=3.( 1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)

3S=3+3^2+3^3+....+3^10

3S-S=3+3^2+3^3+....+3^10-(1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)

2S=3^10-1

S=3^10-1/2

HỌC TỐT NHÉ

Ta có : S = 1 + 3¹ + 3² + .... + 3²⁰¹⁸

=> S - 4 =  1 + 3¹ + 3² + .... + 3²⁰¹⁸ - 4

=> S - 4 = 3² + 3³ + 3⁴ + ..... + 3²⁰¹⁸

=> S - 4 = (3² + 3³ + 3⁴ ) + ...... + (3²⁰¹⁶ + 3²⁰¹⁷ + 3²⁰¹⁸)

=> S - 4 = 3(3 + 3² + 3³) + ..... + 3²⁰¹⁵(3 + 3² + 3³)

=> S - 4 = 3.39 + .... + 3²⁰¹⁵.39

=> S - 4 = 39 (3 + .... + 3²⁰¹⁵) chia hết cho 39

Ta thấy S=(3S-S):2

S=3^0+3^1+3^2+...+3^2018

\(\Rightarrow\)3S=3+3^2+3^3+...+3^2019

\(\Rightarrow\)3S-S=(3+3^2+3^3+..+3^2019)-(3^0+3^1+3^2+...+3^2018)

\(\Rightarrow\)3S-S=3^2019-3^0=3^2019-1\(\Rightarrow\)conf thiếu để bên dưới

S =  1 + 2 + 3 + 4 +.....+ 298 + 299

Dãy số trên là dãy số cách đều với khoảng cách là:

           2 - 1 = 1

Số số hạng của dãy số trên là:

       ( 299 - 1): 1 + 1 = 299

Áp dụng công thức tính tổng của dãy số cách đều ta có tổng S:

S = ( 299 + 1) \(\times\) 299 : 2 

S = 44850 

44850

a) \(...=-\dfrac{1}{4}.\dfrac{4}{17}.\left(-\dfrac{63}{21}\right).\left(-\dfrac{7}{12}\right)\)

\(=-\dfrac{1}{17}.\dfrac{63}{21}.\dfrac{7}{12}\)

\(=-\dfrac{7}{68}\)

b) \(...=-\dfrac{2}{5}.\dfrac{4}{15}-\dfrac{3}{10}.\dfrac{4}{15}\)

\(=\dfrac{4}{15}\left(-\dfrac{2}{5}-\dfrac{3}{10}\right)\)

\(=\dfrac{4}{15}\left(-\dfrac{4}{10}-\dfrac{3}{10}\right)\)

\(=\dfrac{4}{15}.\left(-\dfrac{7}{10}\right)=-\dfrac{14}{75}\)

c) \(...=21-\dfrac{15}{4}:\left(\dfrac{9}{24}-\dfrac{4}{24}\right)\)

\(=21-\dfrac{15}{4}:\dfrac{5}{24}\)

\(=21-\dfrac{15}{4}.\dfrac{24}{5}\)

\(=21-3.6=3\)

d) \(...=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right).\dfrac{7}{3}\)

\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)

\(=\dfrac{7}{3}\left(-\dfrac{3}{4}-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right)\)

\(=\dfrac{7}{3}\left(-1+1\right)=0\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}< 1\)

Chứng tỏ S < 1

Ủng hộ mk nha ^_^

S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)

  \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)

   \(=1-\frac{1}{46}=\frac{45}{46}< 1\)

S=1-1/4 + 1/4 - 1/7 +1/7 - 1/10 +...+ 1/40 - 1/43 +1/43 -1/46

  = 1-1/46 < 1 (ĐPCM)